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Chapter 6. Describing Functions Analysis

Modern Control Theory (Course Code: 10213403)

Professor Jun WANG

( )

Department of Control Science & Engineering

School of Electronic & Information Engineering

Tongji University

Spring semester, 2012


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Outline

1 6.1 Introduction

2 6.2 Calculation of describing functions

3 6.3 Typical describing functions

4 6.4 Stability analysis by describing functions

5 6.5 Limit cycle & its stability

6 6.6 Further discussions

7 6.7 Simulations with MATLAB


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Outline

1 6.1 Introduction








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6.1 Introduction

Introduction

Motivations Frequency response methods based on transfer functions are the

main tool for linear system analyses and syntheses. Can we find

something akin to transfer functions for nonlinear systems?

Observations In response to a sinusoidal signal, most nonlinearities will produce

aperiodic(not necessarily sinusoidal) signal with frequencies being

the harmonics of the input frequency.

Assumptions Approximate the output by the first harmonic alone and neglect the

rest (the nonlinearity behaves as a low-pass filter).

The nonlinearity is time invariant.

There is a single nonlinear element in the system.

Prof J Wang (Tongji Uni)


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6.1 Introduction

Consider a nonlinear elementy =f(u)shown below.

u f(u) y

Letu(t) =Xsin(t),y(t)will be periodic with a fundamental

frequency .

Then,y(t)can be described by the following Fourier series:

y(t) =A0

2 +

n=1

An cos(nt) +Bn sin(nt)

=A0

2 +

n=1 Yn sin(n

t+

n)

where = 2TAn =

1

20 y(t) cos(nt) d(t), Bn =

1

20 y(t) sin(nt) d(t)

Yn =

A2

n+B2

n, n =arctan

An

BnProf J Wang (Tongji Uni) Chap 6. Describing function analysis Spring 2012 5 /57

6 1 I d i


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6.1 Introduction

y(t) =

A0

2 +

n=1

An cos

(n

t) +

Bn sin(n

t)= A0

2 +

n=1 Yn sin(n

t+

n)

y(t)A1 cos(t) +B1 sin(t) =Y1 sin(t+ 1)

The nonlinear element can be described by the first fundamentalcomponent of the above series.

It seems as if functionfwere a linear system with a gain ofY1and

phase of1.

Prof J Wang (Tongji Uni) Chap 6. Describing function analysis Spring 2012 6 /57

6 1 I t d ti


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6.1 Introduction

What is a describing function?

DefinitionThedescribing functionis the complex ratio of the amplitude of the

fundamental component of the output of the nonlinear element to the

sinusoidal input signal.

u f(u) yXsin(t) Y1 sin(t+ 1)

N(X,) = B1+jA1X

=Y1(X,)

X ej1

Y1: Amplitude of the fundamental

harmonic component of the output.

1: Phase shift of the fundamental

harmonic component of the output.Prof J Wang (Tongji Uni) Chap 6. Describing function analysis Spring 2012 7 /57


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Outline

1 6.1 Introduction







6 2 Calculation of describing functions


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6.2 Calculation of describing functions

How to calculate describing functions?

Computation of the describing function is generallystraightforward, but tedious.

It can be computed either analytically or numerically, and may be

determined by experiments.

Some tips for computingy(t)A1 cos(t) +B1 sin(t). Odd functiony(t) = y(t)and even functiony(t) =y(t)

t

y(t)

t

y(t)

The describing functions for odd and memoryless nonlinearities

(saturation, relay, deadzone) are frequency independent (A1 =0).



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Example

Find the describing function of a dead zone and saturation

nonlinearity.

Solutions

(i) Letx(t) =Xsint

NL is symmetric and

momeryless.

= Output is oddfunction and onlyB1is

to be calculated.

x

y

D S

k

t

y

t

x



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(ii) Two important

angles:

d =arcsinD

X

s=

arcsin

S

X

When t(d,s),y=k(xD)

When

t(s, 1/2),y=k(SD)

x

y

D S

k

t

y

DS

t

xDS




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g

t

y

DS

k(SD)(iii) Calculation ofB1

B1 =

1

2

0 y(t) sint d(t)

= 4

2

0y(t) sint d(t)

= 4

s

d

k(XsintD) sint

d(t) +k(SD)

2

ssin(t) d(t)

= omitted

=

2kX

s d

1

2 (sin2ssin 2d) +

2S

X coss

2D

X cosd




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g

sind = DX sin(2d) =2 sindcosd = 2DX 1

DX

2

sins = SX sin(2s) =2 sinscoss = 2SX

1

SX

2

Then

B1 =2kX

s d+

1

2(sin2ssin 2d)

=kX

[2s2d+sin 2ssin 2d]

(iv)

N=

B1

X =

k

[2s

2d

+sin 2s

sin 2d

] =f S

X,

D

X




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Case 1: Pure dead zone

S

( orX< S), s =

2 , sin 2s =0.

N= k

2s2d+sin(2s) sin(2d)

=

k

2dsin(2d)

=k 2k

arcsin DX +

DX

1

DX

2

(XD)

0

1

0 1DX

Nk

Re

Im|

1NX= D X

1k




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Case 2: Pure saturation

D=0, d

=0, sin(2d

) =0

N= k

2s2d+sin(2s) sin(2d)=

k

2s+sin(2s)

=

2k

arcsin SX+

SX

1

SX

2

(XS)

0

1

0 1DX

Nk

Re

Im|

1NX X= S

1k



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Outline

1 6.1 Introduction







6.3 Typical describing functions


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Ideal relay (on-offnonlinearity)

B1 = 1

20

y(t) sin(t) d(t)

=2M

0

sin(t) d(t)

=2M

cos(t)

0

=4M

x

yM

t

y

t

x




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N=4M

X

0

1

0 1MX

N4

1

N =

X

4MRe

Im1

N

X X= 0




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On-offrelay with hysteresis

x(t) =Xsin(t)

y=

M 0 t < t1

t2 < t T

M t1 3 K=3 K< 3

Gp(j) = K

j(1+j)(1+0.5j)

Gp(j) = 0.52 =1, i.e. =

2

Gp(j

2)

=1

K= j2(1+j2)(1+0.5j2)K=3

System is unstable whenK

3There exists a limit cycle

Limit cycle is unstable


6.5 Limit cycle & its stability


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Example (Determination of stability with a hysteresis nonlinearity)

Consider the system with a hysteresis nonlinearity shown below.

Determine whether the system is stable, and find the amplitude andfrequency of the limiting cycle.

R(j) xy

h

MG(s) = 1s(s+1) Y(j)

E(j)



S l ti


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Solutions

The Nyquist plot for the system is shown below.

10

10

20

30

4050

60

70

80

90

100

0.20.40.60.81.0 Re

Im

0.2

0.4

0.6

0.8

1.0

0.20.40.60.81.0 Re

Im



The negative reciprocal of the hysteresis DF is


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The negative reciprocal of the hysteresis DF is

1

N =

1

4MX

1

hX

2

j hX

= 4M

X2 h2 +jhIn this case,M =1 andh =0.1, and we have

1

N =

4

X2

0.01+

j0.1

0.1

0.2

0.10.20.3Re

Im

1N

G(j)

X= h = 0.1 X



The intersection of two curves yields the frequency and the


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The intersection of two curves yields the frequency and the

corresponding amplitude of the stable limit cycle.

1N

= 4

X2 0.01+j0.1

=G(j) = 1

j(j +1)

=2.2 rad/s,X=0.24

0.1

0.2

0.10.20.3Re

Im

1N

G(j)

X= h = 0.1 X

= 2.2X= 0.24



The simulation shows that the limit cycle has an amplitude of 0 24 and


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The simulation shows that the limit cycle has an amplitude of 0.24 and

a frequency 2.2 rad/s.

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

0.25 10 15 20 25 30 35 40 45

t(s)

y



Example


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Example

Determine and analyze the limit cycles of a system described by the

following equation:

x+ x=

1 xx > 0

1 xx < 0

Solutions

(i) Draw block diagram of the CL system

r= 0 eu Gp(s) y(t)xx

e(t)x+ xu(t)

x x



(ii) Calculate the plant transfer function


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(ii) Calculate the plant transfer function

Gp(s) = Y(s)

U(s)

= (1s)X(s)

(s2

+s)X(s)

= 1s

s(s+1)and describing function of the nonlinearity

N(E) =4M

E =

4

E

(iii) Stability analysis

CL system is critically stable

Limit cycle is stable

Re

Im1

N

G(j)

E= 0E 1



(iv) Calculation of frequency and amplitude


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(iv) Calculation of frequency and amplitude

Gp(j) = 1j

j(j +1)

= 2 +j(1 2)

(1+ 2

)

N(E)Gp(j) = 4

E 2 +j(1

2)

(1+ 2) = 1

ReIm1N

G(j)

E= 0E 1

Im

N(E)Gp(j)

=1 2 =0

Re

N(E)Gp(j1)

= 4

E = 1

=1 rad/s

E= 4

=1.2733

(v) The amplitude XSolving the differential equation ofxgivesX= 2

2

.

Hint: x(t) x(t) =e(t) =E sin(t+ )



Time response of x with x(0) = 0 and x(0) = 0


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Time response ofxwithx(0) 0 andx(0) 0

0

0.5

1.0

1.5

0.5

1.0

1.55 10 15 20 25 30 35 40

t(s)

x



Time response of x with x(0) = 0 and x(0) = 0


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Time response ofxwith x(0) 0 and x(0) 0

0

0.5

1.0

1.5

0.5

1.0

1.55 10 15 20 25 30 35 40

t(s)

x


Outline


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1 6.1 Introduction







6.6 Further discussions

Remarks on Describing Functions


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Remarks on Describing Functions

DF method gives the information about stability, but not transient

responses.

How accurate is the DF method?DF method is an approximate

one. The more perpendicularGp(j)is to 1/N, the more

accurate the result.

Gp 1

N

moreaccurate

Gp

1

N

lessaccurate

The above DF method is more accurate for sinusoidal inputs than

for other inputs.The difficulty and accuracy by using DF method depend only on

the complexity of NL components.



For cascade nonlinearities, their describing function is usually


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g y

NOT the multiplicity of individual describing functions.

xM1

M1

M2

M2

zy

xM2

M2

z

xk1

1

1

k1

k2

2

2

k2

zy x

k

k

z

k=k1k2, = 1+

2

k1



If we have a nonlinear element


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c(x) =y(x) +z(x)

N= C1X

=N1+N2

whereN1 = Y1

X andN2 = Z1

X.

x

N1(X)

N2(X)

z

y

+

z +

x

y

k1

x

z

D

k2

x

c=y+z

D

k1+k2


Outline


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1 6.1 Introduction







6.7 Simulations with MATLAB

Simulink Nonlinear Blocks


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With Simulink, we can move beyond idealized linear models to

explore more realistic nonlinear models that describe real-world

phenomena.

Simulink provides a graphical user interface (GUI) for building

models as block diagrams, allowing you to draw models as youwould with pencil and paper.

Models are hierarchical, so you can build models using both

top-down and bottom-up approaches.




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Nonlinear blocks in Simulink


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Backlash Model behavior of system with play

Coulomb and viscous friction Model discontinuity at zero, with linear

gain elsewhere

Dead zone Provide region of zero output

Dead Zone dynamic Set inputs within bounds to zeroHit crossing Detect crossing point

Quantizer Discretize input at specified interval

Rate limiter Limit rate of change of signal

Rate limiter Dynamic Limit rising and falling rates of signalRelay Switch output between two constants

Saturation Limit range of signal

Saturation dynamic Bound range of input

Wra to zero Set out utto zero if in ut is above thresholdProf J Wang (Tongji Uni) Chap 6. Describing function analysis Spring 2012 56 /57



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Modern Control Theory (Course Code: 10213403)

Professor Jun WANG

(

)

Department of Control Science & EngineeringSchool of Electronic & Information Engineering

Tongji University

Spring semester, 2012

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